Question 128: Ray Optics

Question

A ray of light entering from air into a denser medium of refractive index $${4 \over 3}$$, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $$	heta$$ should be equal to :

Accepted Answer

At maximum angle $$	heta$$ ray at point B goes in gazing emergence, at all less values of $$	heta$$, TIR occurs.
At point B
$${4 \over 3} 	imes \sin 	heta '' = 1 	imes \sin 90^\circ $$
$$	heta '' = {\sin ^{ - 1}}\left( {{3 \over 4}} ight)$$
$$	heta ' = \left( {{\pi  \over 2} - 	heta ''} ight)$$
At point A
$$1 	imes \sin 	heta  = {4 \over 3} 	imes \sin 	heta '$$
$$\sin 	heta  = {4 \over 3} 	imes \sin \left( {{\pi  \over 2} - 	heta ''} ight)$$
$$\sin 	heta  = {4 \over 3}\cos \left[ {{{\cos }^{ - 1}}{{\sqrt 7 } \over 4}} ight]$$
$$\sin 	heta  = {4 \over 3} 	imes {{\sqrt 7 } \over 4}$$
$$	heta  = {\sin ^{ - 1}}\left( {{{\sqrt 7 } \over 3}} ight)$$

Answer

$${\sin ^{ - 1}}{{\sqrt 7 } \over 3}$$

Answer

$${\sin ^{ - 1}}{{\sqrt 5 } \over 4}$$

Answer

$${\sin ^{ - 1}}{{\sqrt 7 } \over 4}$$

Answer

$${\sin ^{ - 1}}{{\sqrt 5 } \over 3}$$

Ray Optics

A ray of light entering from air into a denser medium of refractive index ${4 \over 3}, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle \theta$ should be equal to :

Choose the correct answer: